Idempotent Ring is Commutative

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Let $\struct {R, +, \circ}$ be an idempotent ring.

Denote with $0_R$ the zero of $R$.

Then $\struct {R, +, \circ}$ is a commutative ring.


Let $x, y \in R$.


\(\ds x + y\) \(=\) \(\ds \paren {x + y}^2\) Definition of Idempotent Ring
\(\ds \) \(=\) \(\ds x^2 + x \circ y + y \circ x + y^2\) Binomial Theorem: Ring Theory
\(\ds \) \(=\) \(\ds x + x \circ y + y \circ x + y\) Definition of Idempotent Ring

Subtracting $x + y$ from both sides yields:

\(\ds x \circ y + y \circ x\) \(=\) \(\ds 0_R\)
\(\ds \) \(=\) \(\ds y \circ x + y \circ x\) Idempotent Ring has Characteristic Two

Finally, subtracting $y \circ x$ from both sides, we obtain:

$x \circ y = y \circ x$

and conclude $R$ is a commutative ring.